Question: $h(x) = 4x+2$ $g(n) = n^{3}+2n^{2}-7n+5+f(n)$ $f(t) = 3t-h(t)$ $ g(f(-1)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = (3)(-1)-h(-1)$ To solve for the value of $f$ , we need to solve for the value of $h(-1)$ $h(-1) = (4)(-1)+2$ $h(-1) = -2$ That means $f(-1) = (3)(-1)-(-2)$ $f(-1) = -1$ Now we know that $f(-1) = -1$ . Let's solve for $g(f(-1))$ , which is $g(-1)$ $g(-1) = (-1)^{3}+2(-1)^{2}+(-7)(-1)+5+f(-1)$ To solve for the value of $g$ , we need to solve for the value of $f(-1)$ $f(-1) = (3)(-1)-h(-1)$ To solve for the value of $f$ , we need to solve for the value of $h(-1)$ $h(-1) = (4)(-1)+2$ $h(-1) = -2$ That means $f(-1) = (3)(-1)-(-2)$ $f(-1) = -1$ That means $g(-1) = (-1)^{3}+2(-1)^{2}+(-7)(-1)+5-1$ $g(-1) = 12$